1. Differentials:Please to if my work is correct

I just want to see if my work is correct.

This problems comes in for parts. I am concerned about the last three. I think the processes should be longer but I don't see how I can make them longer.

4.8 #1
y`` + y = 0

a.

e^xsolve with: e^x
e^x y`` + y = 0

a y"+b y' +c y=0

a y’’ +b y’ + c =0

a e^(y’’ x) + b e^(y’ x)

e^(y)(a cos(x) +b sin(x))

This is not a solution

b.Sin x
sin x y'' + y = 0

1/2 i e^-i^x y``(x)- 1/2 i e^i^x y``(x)+y(x)=0
y(x)+sin(x) = cos(x) y''(x)
Solution

c.Cos x
cos x y'' + y = 0

1/2 e^(-i x) y''(x)+1/2 e^(i x) y''(x)+y(x) = 0cos(x) y''(x)+y(x) = 0Solution

d.Sin x – cos x

-1/2 e^(-i x) y''(x)-1/2 e^(i x) y''(x)+y(x)+1/2 i e^(-i x)-1/2 i e^(i x) = 0

Solution

2. Originally Posted by pikachumanson
I just want to see if my work is correct.

This problems comes in for parts. I am concerned about the last three. I think the processes should be longer but I don't see how I can make them longer.

4.8 #1
y`` + y = 0

a.

e^xsolve with: e^x
e^x y`` + y = 0

a y"+b y' +c y=0

a y’’ +b y’ + c =0

a e^(y’’ x) + b e^(y’ x)

e^(y)(a cos(x) +b sin(x))

This is not a solution

b.Sin x
sin x y'' + y = 0

1/2 i e^-i^x y``(x)- 1/2 i e^i^x y``(x)+y(x)=0
y(x)+sin(x) = cos(x) y''(x)
Solution

c.Cos x
cos x y'' + y = 0

1/2 e^(-i x) y''(x)+1/2 e^(i x) y''(x)+y(x) = 0cos(x) y''(x)+y(x) = 0Solution

d.Sin x – cos x

-1/2 e^(-i x) y''(x)-1/2 e^(i x) y''(x)+y(x)+1/2 i e^(-i x)-1/2 i e^(i x) = 0

Solution
Sorry but I can't make any sense of this. Please post the questions exactly as they are written from wherever you got them from.

3. [QUOTE=pikachumanson;533340]I just want to see if my work is correct.

This problems comes in for parts. I am concerned about the last three. I think the processes should be longer but I don't see how I can make them longer.

4.8 #1
y`` + y = 0
Is this the differential equation you want to solve?

[quote[ a.

e^xsolve with: e^x
What does that mean?

e^x y`` + y = 0
Just put $e^x$ in front of y''? Why?

a y"+b y' +c y=0

a y’’ +b y’ + c =0

a e^(y’’ x) + b e^(y’ x)
Now you have taken the exponential of each derivative. Again, why? And where did a, b, and c suddenly come from?

e^(y)(a cos(x) +b sin(x))

This is not a solution

b.Sin x
sin x y'' + y = 0

1/2 i e^-i^x y``(x)- 1/2 i e^i^x y``(x)+y(x)=0
y(x)+sin(x) = cos(x) y''(x)
Solution

c.Cos x
cos x y'' + y = 0

1/2 e^(-i x) y''(x)+1/2 e^(i x) y''(x)+y(x) = 0cos(x) y''(x)+y(x) = 0Solution

d.Sin x – cos x

-1/2 e^(-i x) y''(x)-1/2 e^(i x) y''(x)+y(x)+1/2 i e^(-i x)-1/2 i e^(i x) = 0

Solution
I agree with mr. fantastic- what you wrote makes no sense at all!